Optical instruments are how we see the world, from corrective eyewear to medical endoscopes to cell phone cameras to orbiting telescopes. When you finish this course, you will be able to design, to first order, such optical systems with simple mathematical and graphical techniques. This first order design will allow you to develop the foundation needed to begin all optical design as well as the intuition needed to quickly address the feasibility of complicated designs during brainstorming meetings. You will learn how to enter these designs into an industry-standard design tool, OpticStudio by Zemax, to analyze and improve performance with powerful automatic optimization methods.

AY

Learned a ton about paraxial optics design. Great lecture, capstone project is A LOT of work (>20 hour) for me.

FH

Apr 14, 2019

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This is a good course for those people wanting to start their learning in optical system design.

From the lesson

First Order Ray Tracing

This module applies Snell's Laws to the curved interfaces of lenses and mirrors. You will learn the graphical and mathematical tools you need to calculate image size and location for thin lens and mirror systems. This foundation is essential to build more complex optical systems.

Taught By

Amy Sullivan

Research Associate

Robert McLeod

Professor

Transcript

Let's look at a laboratory a demonstration of the gray diagrams that we've been talking about thus far. So in this case, they have a light source that has three rays that are entering this white paper, and these are just three laser beams that are designed to represent three rays that we were talking about in our ray diagrams. Now, we'd like to see what happens when we use this light source with the number of lenses. A positive, two positive lenses, and a negative lens. So, three parallel rays appear to be coming from a light source from an object that is infinitely far away, and so the rays are parallel. So think about the situation, where we have an object infinitely far away then we send those rays through a positive lens. What do you think is going to happen on the other side? Let's go ahead and try it, that's why we've got toys to play with. So if I put these parallel race into its positive lens, they will converge, form an object that is at the back focal plane of this positive lens. This is one focal length away from our lens. This turns out to actually be a pretty nice way of finding a focal length for positive lens. If you have a source that you can have parallel rays, you can measure where they converge and that will give you your focal length. How about a negative lens? So we got a positive lens first, let's try a negative lens. What do you think will happen when we put the negative lens in these parallel rays? Let's go ahead and do that. All right, now do not have a clear image. I do not have any converging rays on the opposite side of this lens. So my rays are all diverging. But if I carefully trace them back through the lens, I will notice that all of these rays will overlap at the front focal length of this lens. And so, this also gives me the front focal plane, this gives me the focal length of a negative lens, and instead of a real image out here, which I can see where the rays are converging, I have the virtual image in front of the lens where the rays appear to be coming from. Now let's try a couple of different lenses. So, I've already tried this positive lens, and it looks like it has a focal length. How about this, let me put my pencil here, cause we have a rough idea of where it was. Now, I have another positive lens, I would like to see about were that focal length is, and it looks like, I line these up roughly, that is just focusing a little earlier. So now the parallel rays that are converging at the front focal plane where I see my image is a little sooner, so I have a focal length that is shorter. So what happens when I take this image, and use this as the object for a second lens? So now instead of parallel rays coming in, I have three rays that are all coming from a point. So what will happen now, when I put this positive lens into this set of rays? I think in this case, it matters where I put the lens. Trying t put the lens about one focal length away, I think that's about right, and the reason I know that's about right is if all of my rays are coming from the front focal plane, then I know they will exit parallel, and I see they're approximately parallel here, so this must be one focal length away. This situation is the a focal situation, where I have parallel rays coming in, and I have parallel rays coming out. Now, I have an object, that is a real object, and my image is infinitely far away. I have parallel rays leaving, and so I have an image that's infinitely far away, and so this is my infinite conjugate case. My conjugate planes, the object plane, and the image plane, are infinitely far away from each other. Let's make them less infinitely far away. Let's put them closer together and form a real image. Do you think I should move the positive lens towards the outer edge or ashamedly farther away? Well, why don't we go ahead and try. Let's first try moving it towards. All right. As I move towards this object, now the object is closer to this lens than the focal length, and now my rays are diverging. And so, I don't have a real image over here. I have a virtual image, where those rays appear to be coming from, that is in the front of the lens, and I could draw my rays backwards, and I could find that. What if I move farther away? So, now, I go through the point where I have parallel rays exiting, and I keep moving farther away. At some point, I start to see an image here, and so, I have an object plane in front of the lens, and I have an image plane on the far side of the lens. These skins are our conjugate planes, our object plane, and our image plane. And I can find that using this object distance, then I can find my image distance. Well, two positive lenses were fun. Let's see if we can try a negative lens and a positive lens. I'm going to try the negative lens first. I'm going to put this thing here. And so now I have a- What do I have? I've got an object that's infinitely far away, and an image, where is my image? Right. So, I have an image, it's a virtual image. It's going to be in front of this lens, it's a negative distance away. So what if I take this image, where are these rays converge back here, in this virtual image, and use that as my object instead, for this positive lens? Well, we're going to take that object, and this lens will form an image. So even if I have a virtual object, I can form a real image with an additional lens in my system.

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